What is the probability density of $R = \sin(X) * \cos(Y)$ where $X,Y$ are independent random variables, uniformly distributed on $[0, 2 \pi]$?
I am stuck with complicated integrals, not sure if there is a closed expression for the density. Can anybody help me, please?
$\sin X$ and $\cos Y$ are identically distributed, with $\Pr(\sin X\le x)=\Pr(X\le\sin^{-1}(x))$ so the density of $\sin X$ is $\frac{d}{dx}\sin^{-1}(x)=\frac1{\sqrt{1-x^2}}$. Thus the product distribution is, for $-1\le z\le 1$, $$ f(z)=\int_{-1}^1\frac1{\sqrt{1-x^2}}\frac{1}{\sqrt{1-(z/x)^2}}\frac1{|x|}\,1_{[-x,x]}(z)\,dx $$ $$=\left(\int_{-1}^{-|z|}+\int_{|z|}^1\right) \frac1{\sqrt{1-x^2}}\frac{|x|}{\sqrt{x^2-z^2}}\frac1{|x|}\,dx $$ $$=\left(\int_{-1}^{-|z|}+\int_{|z|}^1\right) \frac1{\sqrt{1-x^2}}\frac{1}{\sqrt{x^2-z^2}}\,dx $$ $$=2\int_{|z|}^1 \frac1{\sqrt{1-x^2}}\frac{1}{\sqrt{x^2-z^2}}\,dx $$ $$ =2 K(1-z^2) $$ where $K(m)$ is the complete elliptic integral of the first kind with parameter $m=k^2$ (source: Wolfram Alpha for the last part).
